Question: Let $f(x) = -8x^{2}+8x+6$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-8x^{2}+8x+6 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -8, b = 8, c = 6$ $ x = \dfrac{-8 \pm \sqrt{8^{2} - 4 \cdot -8 \cdot 6}}{2 \cdot -8}$ $ x = \dfrac{-8 \pm \sqrt{256}}{-16}$ $ x = \dfrac{-8 \pm 16}{-16}$ $x =-\frac{1}{2},\frac{3}{2}$